Step-by-step Solution

1. Define the Coordinate System and Motion Let's set up a coordinate system. Let the center of the circular path, O, be the origin (0, 0, 0). The motion of the mass m occurs in the xy-plane. The point P, where the string is fixed, is on the z-axis at a height h above the origin. So, the coordinates of P are (0, 0, h).

   The mass m undergoes uniform circular motion with a constant angular speed ω in a circle of radius r. The position vector of the mass at any time t can be written as: $\vec{r} = r \cos(\omega t) \hat{i} + r \sin(\omega t) \hat{j}$ The velocity vector is the time derivative of the position vector: $\vec{v} = \frac{d\vec{r}}{dt} = -r\omega \sin(\omega t) \hat{i} + r\omega \cos(\omega t) \hat{j}$ The linear momentum of the mass is $\vec{p} = m\vec{v}$: $\vec{p} = -mr\omega \sin(\omega t) \hat{i} + mr\omega \cos(\omega t) \hat{j}$

2. Calculate Angular Momentum about O (${\overrightarrow L _O}$) The angular momentum about the origin O is given by the formula ${\overrightarrow L _O} = \vec{r} \times \vec{p}$. ${\overrightarrow L _O} = (r \cos(\omega t) \hat{i} + r \sin(\omega t) \hat{j}) \times (-mr\omega \sin(\omega t) \hat{i} + mr\omega \cos(\omega t) \hat{j})$ We perform the cross product, recalling that $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{i} = -\hat{k}$: ${\overrightarrow L _O} = (r \cos(\omega t))(mr\omega \cos(\omega t)) (\hat{i} \times \hat{j}) + (r \sin(\omega t))(-mr\omega \sin(\omega t)) (\hat{j} \times \hat{i})$ ${\overrightarrow L _O} = mr^2\omega \cos^2(\omega t) \hat{k} - mr^2\omega \sin^2(\omega t) (-\hat{k})$ ${\overrightarrow L _O} = mr^2\omega \cos^2(\omega t) \hat{k} + mr^2\omega \sin^2(\omega t) \hat{k}$ Using the trigonometric identity $\cos^2\theta + \sin^2\theta = 1$: ${\overrightarrow L _O} = mr^2\omega (\cos^2(\omega t) + \sin^2(\omega t)) \hat{k} = mr^2\omega \hat{k}$ The resulting vector ${\overrightarrow L _O}$ has a constant magnitude ($mr^2\omega$) and a constant direction (along the positive z-axis). Therefore, ${\overrightarrow L _O}$ remains constant with time.

   Alternative method (Torque): The net force on the mass for uniform circular motion is the centripetal force, which is always directed towards the center O. So, $\vec{F}_{net}$ is parallel to $-\vec{r}$. The torque about O is $\vec{\tau}_O = \vec{r} \times \vec{F}_{net}$. Since $\vec{r}$ and $\vec{F}_{net}$ are anti-parallel, their cross product is zero. $\vec{\tau}_O = 0$. Since $\vec{\tau}_O = \frac{d{\overrightarrow L _O}}{dt}$, we have $\frac{d{\overrightarrow L _O}}{dt} = 0$, which means ${\overrightarrow L _O}$ is constant.

3. Calculate Angular Momentum about P (${\overrightarrow L _P}$) The position of point P is $\vec{r}_P = h\hat{k}$. The position vector of the mass m relative to P is $\vec{r}_{m/P} = \vec{r} - \vec{r}_P$. $\vec{r}_{m/P} = (r \cos(\omega t) \hat{i} + r \sin(\omega t) \hat{j}) - h\hat{k}$ The angular momentum about P is given by ${\overrightarrow L _P} = \vec{r}_{m/P} \times \vec{p}$. ${\overrightarrow L _P} = ((r \cos(\omega t) \hat{i} + r \sin(\omega t) \hat{j}) - h\hat{k}) \times (-mr\omega \sin(\omega t) \hat{i} + mr\omega \cos(\omega t) \hat{j})$ We can split this into two parts: ${\overrightarrow L _P} = (r \cos(\omega t) \hat{i} + r \sin(\omega t) \hat{j}) \times \vec{p} + (-h\hat{k}) \times \vec{p}$ The first term is exactly ${\overrightarrow L _O}$, which we found to be $mr^2\omega \hat{k}$. Let's calculate the second term: $(-h\hat{k}) \times (-mr\omega \sin(\omega t) \hat{i} + mr\omega \cos(\omega t) \hat{j})$ $= (-h)(-mr\omega \sin(\omega t)) (\hat{k} \times \hat{i}) + (-h)(mr\omega \cos(\omega t)) (\hat{k} \times \hat{j})$ Using $\hat{k} \times \hat{i} = \hat{j}$ and $\hat{k} \times \hat{j} = -\hat{i}$: $= hmr\omega \sin(\omega t) \hat{j} - hmr\omega \cos(\omega t) (-\hat{i})$ $= hmr\omega \cos(\omega t) \hat{i} + hmr\omega \sin(\omega t) \hat{j}$ Now, we combine the two parts to get ${\overrightarrow L _P}$: ${\overrightarrow L _P} = mr^2\omega \hat{k} + hmr\omega \cos(\omega t) \hat{i} + hmr\omega \sin(\omega t) \hat{j}$ The vector ${\overrightarrow L _P}$ has a constant z-component, but its x and y components, $hmr\omega \cos(\omega t)$ and $hmr\omega \sin(\omega t)$, vary with time. Since the components of the vector are changing, the vector ${\overrightarrow L _P}$ itself varies with time. Specifically, its direction changes, precessing around the z-axis.

4. Conclusion We have found that ${\overrightarrow L _O}$ is a constant vector, while ${\overrightarrow L _P}$ is a time-varying vector.

   • ${\overrightarrow L _O}$ remains constant.
   • ${\overrightarrow L _P}$ varies with time.

   Comparing this conclusion with the given options: A: ${\overrightarrow L _O}$ and ${\overrightarrow L _P}$ do not vary with time. (Incorrect) B: ${\overrightarrow L _O}$ varies with time while ${\overrightarrow L _P}$ remains constant. (Incorrect) C: ${\overrightarrow L _O}$ remains constant while ${\overrightarrow L _P}$ varies with time. (Correct) D: ${\overrightarrow L _O}$ and ${\overrightarrow L _P}$ both vary with time. (Incorrect)